Assume that \(X\cong P\Omega _{n}^{+}(q)\)  with \(n \geqslant 4\) . Actually, \(n\geqslant 8\)  by Remark REF  and since \(P\Omega _{6}^{+}(q)\cong PSL_{4}(q)\)  is ruled out in Lemma REF . The group \(X_{x}\)  lies in a large maximal geometric subgroup \(Y\)  of \(X\)  by
Theorem REF . Then \(Y\)  is one of the the groups listed in [1]}, Proposition 4.23. Filtering such a list with respect to the constraint \(\left( \Phi _{(n-2)f}^{\ast }(p),\left|Y\right|\right) >1\) , we see
that either \(Y\in \mathcal {C}_{1}(X)\) , or \(Y\)  is a \(\mathcal {C}_{3}\) -group
of type \(O_{n/2}(q^2)\)  or \(GU_{n/2}(q)\)  according to whether \(n/2\)  is odd or even, respectively, or \((X,Y)=(P\Omega _{8}^{+}(3),O_{1}(3)\wr S_{8}),(P\Omega _{8}^{+}(3),2^{6}.\Omega _{6}^{+}(2))\) . The latter is immediately ruled out.
Indeed, \(13\mid \left|X\right|\)  but \(13^{2}\nmid \left|X\right|\)  and \(13 \nmid \left|Y\right|\) .
Assume that \(Y\)  is a maximal \(\mathcal {C}_{3}\) -subgroup of \(X\)  of type \(O_{n/2}(q^2)\) . Then \(X\)  has one or two conjugacy classes of subgroups
isomorphic to \(Y\)  by [2]}, Propositions 4.3.20(I) and 2.5.10(i), according to whether \(q \equiv 3 \pmod {4}\)  or \(q \equiv 1 \pmod {4}\)  respectively. In the former case \(X_{x}=Y\)  by Lemma REF (1), whereas in the latter \(G_{x}\)  is a type 1 novelty with respect to \(Y\)  by Lemma REF (2). However, this is impossible by [2]}, Tables 3.5.H–I. Thus \(X_{x}=Y\)  in each case, and hence
\(k^{2}=q^{\left( n/2-1\right) ^{2}}\frac{(q-1)^{2}(q+1)}{(4,q^{n/2}-1)}\cdot \frac{q^{n/2}-1}{q-1}\) 
by [2]}, Proposition 4.3.20.(II). Then \(\frac{q^{n/2}-1}{q-1}\)  is a square, being coprime to the other factor of \(\frac{4k^{2}}{(q+1)^{2}}\)  or \(\frac{k^{2}}{(q+1)^{2}}\)  according to whether \(q \equiv 1 \pmod {4}\)  or \(q \equiv 3 \pmod {4}\)  respectively. Hence \((n,q)=(10,3)\)  by [5]}, A8.1, since \(n/2\)  is odd. However, such a pair does not fulfill (REF ), and hence this case is excluded.
Assume that \(Y\)  is a maximal \(\mathcal {C}_{3}\) -subgroup of \(X\)  of type \(GU_{n/2}(q)\)  with \(n/2\)  even. Then \(X\)  has two conjugacy classes of subgroups
isomorphic to \(Y\)  by [2]}, Proposition 4.3.18.(I). If \(X_{x}<Y\) , then \(G_{x}\)  is a type 1 novelty with respect to \(Y\)  by Lemma REF (2).
However, this is impossible by [2]}, Tables 3.5.H–I. Thus \(X_{x}=Y\) 
and hence
\(k^{2}=\frac{1}{2} q^{n\left( n-2\right) /8} \prod _{i=1}^{n/2-1}(q^{i}+(-1)^{i})\) 
Arguing as in the \(P\Omega ^{-}(q)\) -case, and bearing in mind that \(n/2\)  is even, we see that
\(k+1 \mid f(q+1,n/2)(q+1,n/2-2)(q^{n/2}-1)(q^{n/2-1}+1)\) 
and hence
\(q^{n(n-2)/4-1 }/2<k^{2}<q^{2}(q+1)^{4}(q^{n/2}+1)^{2}(q^{n/2-1}-1)^{2}<64q^{2n+4},\) 
which yields either \(n=8\) , or \(n=12\)  and \(q<64\) . It is easy to check that, there are no \(q\)  less than 64 fulfilling REF  for \(n=12\) . Thus \(n=8\)  and hence (REF ) becomes
\(k^{2}=\frac{1}{2}q^{6}\left( q^{2}+1\right) (q-1)^{2}\left( q^{2}+q+1\right).\) 
So \(q^{2}+q+1\)  is a square. However, this is impossible by [5]}, A7.1. Thus \(Y\)  is not a \(\mathcal {C}_{3}\) -subgroup of \(X\) .
Assume that \(Y\in \mathcal {C}_{1}(X)\) . Then \(Y\)  is not a parabolic
subgroup of \(X\)  by [2]}, Proposition 4.1.20.(II), since \(\left( \Phi _{(n-2)f}^{\ast }(p),\left|Y\right|\right) >1\) . Moreover, by [2]}, Propositions 4.1.26.(II)–4.1.7.(II) and 2.5.10.(i), one of the following
cases occurs:

\(Y\)  is the stabilizer in \(X\)  of a non-singular point of \(PG_{n-1}(q)\) :

\(Y\cong \Omega _{n-1}(q)\)  with \(q\equiv 1 \pmod {4}\) , or \(q\equiv 3 \pmod {4}\)  and \(n/2\)  even;

\(Y\cong \Omega _{n-1}(q).Z_{2}\) , with \(q\equiv 3 \pmod {4}\)  and \(n/2\) 
odd;

\(Y\cong Sp_{n-2}(q)\) , with \(q\)  even.


\(Y\)  is the stabilizer in \(X\)  of a non-singular line of type \(-\)  of \(PG_{n-1}(q)\) :

\(Y\cong \left( Z_{\frac{q+1}{(q+1,2)}}\times \Omega _{n-2}^{-}(q)\right) .Z_{2}\) , with \(q\)  even, or \(q\equiv 1 \pmod {4} \) ;

\(Y\cong \left( Z_{\frac{q+1}{2}}\times \Omega _{n-2}^{-}(q)\right).[4] \) , with \(q\equiv 3 \pmod {4}\)  and \(n/2\)  odd;

\(Y\cong Z_{2}. \left( Z_{\frac{q+1}{4}}\times P\Omega _{n-2}^{-}(q)\right) .[4]\) ,
with \(q\equiv 3 \pmod {4}\)  and \(n/2\)  even.


The group \(X\)  has a unique conjugacy class of subgroups isomorphic to \(Y\)  in
case (1c) and in cases (2a)–(2c) by [2]}, Propositions
4.1.6.(I)–4.1.7.(I). Then \(X_{x}=Y\)  in these cases by Lemma REF (1).
In cases (1a) and (1b), \(X\)  has two conjugacy classes of subgroups
isomorphic to \(Y\)  again by [2]}, Proposition 4.1.6.(I). If \(X_{x}<Y\) ,
then \(G_{x}\)  is a type 1 novelty with respect to \(Y\)  by Lemma REF (2),
but this is impossible by [2]} Tables 3.5.H–I. Thus, \(Y=X_{x}\)  in each
case.
Assume that \(X_{x}\)  is the stabilizer in \(X\)  of a non-singular point of \(PG_{n-1}(q)\) . As pointed in [14]}, Section 5 and (1.b), the lengths of
the \(X_{x}\) -orbits distinct from \(\left\lbrace  x\right\rbrace  \)  are as follows:
\(q\equiv 1 \pmod {4}\)
one orbit of length \(q^{n-2}-1\) , one of length \(q^{n/2-1}(q^{n/2-1}+1)/2\) , \((q-1)/4\)  ones of length \(q^{n/2-1}(q^{n/2-1}-1)\) , and
\((q-5)/4\)  ones of length \(q^{n/2-1}(q^{n/2-1}+1)\) .
\(q\equiv 3 \pmod {4}\)
one orbit of length \(q^{n-2}-1\) , one of length \(q^{n/2-1}(q^{n/2-1}-1)/2\) , \((q-3)/4\)  ones of length \(q^{n/2-1}(q^{n/2-1}-1)\) , and
\((q-3)/4\)  others of length \(q^{n/2-1}(q^{n/2-1}+1)\)
\(q \equiv 0 \pmod {2}\)
one of length \(q^{n-2}-1\) , \(q/2\)  ones of length \(q^{n/2-1}(q^{n/2-1}-1)\)  and \((q-2)/2\)  ones of length \(q^{n/2-1}(q^{n/2-1}+1)\) .

In each of the three cases the greatest common divisor of the lengths of the \(X_{x}\) -orbits is 1. Thus \(k+1\mid (2,q-1)\left|\mathrm {Out(}X\mathrm {)}\right|\) , since \(\frac{k+1}{\left( k+1,\left|\mathrm {Out(}X\mathrm {)}\right|\right) }\)  divides the length of each \(X_{x}\) -orbit distinct from \({\lbrace x \rbrace }\)  by Lemma REF (2). Hence, \(k<(2,q-1)\cdot (4,q^{n/2}-1)\cdot 2\theta f\) , where
either \(\theta =3\)  and \(n=8\) , or \(\theta =1\) . Since \(k^{2}=[X:X_{x}]\) , it follows that \(k^{2}_{p}=q^{(n-2)/2}\) , hence \(q^{(n-2)/4}\mid k\) . Thus \(q^{(n-2)/4}<(2,q-1)\cdot (4,q^{n/2}-1)\cdot 2\theta f\)  and hence either \(n=8\)  and \(q=4,9\) , or \(n=10\)  and \(q=3\) . Therefore, either \(X\cong P\Omega _{8}^{+}(4)\)  and \(X_{x}\cong Sp_{6}(2)\) , or \(X \cong P\Omega _{8}^{+}(9)\)  and
\(X_{x} \cong \Omega _{7}(9)\) , or \(X \cong P \Omega _{10}^{+}(3)\)  and \(X_{x} \cong \Omega _{9}(3).Z_{2}\) . The first two cases are ruled out, since \(\left[ X:X_{x}\right] \)  is not a square for any of them, whereas in the latter \(k=99\) . Then \(b=9900\lambda \) , where \(\lambda \mid 99\)  and \(\lambda \geqslant 2\) . Thus \(3^{16}\cdot 7 \cdot 13 \cdot 41 \mid \left|X_{B} \right|\)  and hence \(\Omega _{9}(3) \unlhd X_{B} \leqslant \Omega _{9}(3).Z_{2}\)  by [15]}, Tables 8.58–8.59 and 8.66–8.67. So, either \([X:X_{B}]=99^2\)  or \([X:X_{B}]=2*99^2\) , and hence \(b \mid 8*99^{2}\)  since \(\left|\mathrm {Out(}X\mathrm {)}\right|=4\) . However, this is impossible since 100 divides \(b\) . So, this case is ruled out.
Finally, assume that \(X_{x}\)  is the stabilizer in \(X\)  of a non-singular line of type \(-\)  of \(PG_{n-1}(q)\) . Then \(\left|X_{x}\right|=\left|\Omega _{n-2}^{-}(q)\right|(q+1)\mu \) ,
where
\(\centering \mu = \left\lbrace \begin{split}2 & \hphantom{ciao}q\text{ even, or }q \equiv 3 \hspace{-7.11317pt} \pmod {4}\text{ and } n/2 \text{ odd} \\1/2 & \hphantom{ciao}q \equiv 3 \hspace{-7.11317pt} \pmod {4}\text{ and } n/2 \text{ even} \\1 & \hphantom{ciao}q \equiv 1 \hspace{-7.11317pt} \pmod {4} \text{.}\\\end{split}\right.\) 
Set \(\rho =\frac{\left( q^{n/2-1}+1,4\right) }{\left( q^{n/2}-1,4\right)\mu }\) . Then \(\rho \)  is 2 for \(q \equiv 3 \pmod {4}\)  and \(n/2\)  even, and \(1/2\)  otherwise, and hence
\(k^{2}=q^{n-2}\rho (q-1)^{2}\left[ \frac{1}{(q+1)}\left( \frac{q^{n/2}-1}{q-1}\right) \left( \frac{q^{n/2-1}-1}{q-1}\right) \right] \text{.} \) 
Assume that \(n/2\)  is odd. Then \(\frac{q^{n/2}-1}{q-1}\)  is odd and \(\frac{q^{n/2-1}-1}{q^{2}-1}\)  is an integer. Moreover, they are coprime. Then \(\frac{q^{n/2}-1}{q-1}\)  is a square, since the maximal odd divisor of \(\frac{(q-1)^{2}}{\left( 4,q^{n/2}-1\right) }\)  is a square too. Then \((n,q)=(10,3)\) 
by [5]}, A7.1 and A8.1, since \(n\geqslant 8\) . However, such \((n,q)\)  does
not fulfill (REF ).
Assume that \(n/2\)  is even. Arguing as above, we see that \(\frac{q^{n/2-1}-1}{q-1}\)  is a square and hence \((n,q)=(12,3)\)  by [5]}, A7.1 and A8.1,
since \(n\geqslant 8\) , but such \((n,q)\)  does not fulfill (REF ).
This completes the proof.
